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A ball is shot from the ground into the air. At a height of 7.20 m, its velocity is $\vec{v}$ = (7.00$\hat{i}$ + 3.00$\hat{j}$) m/s, with $\hat{i}$ horizontal and $\hat{j}$ upward. What is the magnitude of the ball's velocity just before it hits the ground?

Sagot :

MathPhys

Answer:

14.1 m/s

Explanation:

The ball is a projectile in free fall, meaning the only force acting on it is the force of gravity. The horizontal and vertical components of the ball's motion can be modeled using constant acceleration equations, also known as SUVAT equations. In this case, we will use the equation:

v² = u² + 2as

where

  • s is the displacement (distance and direction)
  • u is the initial velocity (speed and direction)
  • v is the final velocity (speed and direction)
  • a is the acceleration

In the horizontal direction, there is no acceleration (a = 0). Therefore, the horizontal component of the final velocity is:

v² = u² + 2as

v² = (7.00 m/s)² + 2 (0 m/s²) s

v = 7.00 m/s

In the vertical direction, the displacement is s = -7.20 m, and the acceleration is a = -9.8 m/s². The vertical component of the final velocity is therefore:

v² = u² + 2as

v² = (3.00 m/s)² + 2 (-9.8 m/s²) (-7.20 m)

v = -12.3 m/s

Using Pythagorean theorem, the magnitude of the ball's velocity is:

v² = (7.00 m/s)² + (-12.3 m/s)²

v = 14.1 m/s

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